Simplify the following functional expressions using boolean

Simplify the following functional expressions using boolean algebra and its identities. List the identity used at each step. a) y(xz\' + x\'z) + y\'(xz\' + x\'z) b) x(y\'z +y) + x\'(y +z\')\' c) x[y\'z + (y + z\')\'](x\'y +z)

Solution

a) y(xz\' + x\'z) + y\'(xz\' + x\'z)

Answer :->

y(xz\' + x\'z) + y\'(xz\' + x\'z) =

taking (xz\' + x\'z) common,

= (xz\' + x\'z)(y + y\')

note that (y + y\') = 1 (Complement)

so, (xz\' + x\'z)(y + y\') = (xz\' + x\'z)(1)

= (xz\' + x\'z)

Note that (x xor z) = (xz\' + x\'z),

Verify this from following truth table,

So,

y(xz\' + x\'z) + y\'(xz\' + x\'z) = (x xor z)

b) x(y\'z +y) + x\' (y +z\')\'

Answer :->

x(y\'z +y) + x\' (y +z\')\' =

put (y +z\')\' = y\'z (de Morgan’s Theorem)

= x(y\'z +y) + x\'y\'z

= xy\'z + xy + x\'y\'z =  xy\'z + x\'y\'z  + xy = y\'z(x + x\') + xy,

so using complement, (x + x\') = 1

= y\'z + xy

So,

x(y\'z +y) + x\' (y +z\')\' = = y\'z + xy

c) x[y\'z + (y + z\')\'](x\'y +z)

Answer :->

x[y\'z + (y + z\')\'](x\'y +z)=

note that (y + z\')\' = y\'z (de Morgan’s Theorem)

= x[y\'z + y\'z](x\'y +z)

using idempotent law,

y\'z + y\'z = y\'z

So,

x[y\'z + (y + z\')\'](x\'y +z) = x[y\'z](x\'y +z)